Nonsymmetric conical upper density and $k$-porosity

0.2 MB·

by  Antti Käenmäki

NONSYMMETRIC CONICAL UPPER DENSITY AND
k-POROSITY
ANTTI KA¨ENMA¨KI AND VILLE SUOMALA
7
Abstract. WestudyhowtheHausdorffmeasureisdistributedinnonsymmet-
1 ricnarrowconesinRn. Asanapplication,wefindanupperboundcloseton k
20 for the Hausdorff dimension of sets with large k-porosity. With k-porous s−ets
we mean sets which have holes in k different directions on every small scale.
n
a
J

0
3
1. Introduction

A] It is a well known fact that for a set A Rn with finite s-dimensional Hausdorff
⊂
C measure, s(A) < , we have
H ∞
.
h s A B(x,r)
t 1 limsup H ∩ 2s (1.1)
a ≤ rs ≤
m r 0 (cid:0) (cid:1)
↓
[ for s-almost every x A. For a proof, see, for example, [12, Theorem 6.2(1)].
H ∈
This is analogous to the classical Lebesgue Density Theorem. Using this fact, we
1
v know roughly how much of A there is in small balls. Mattila [11] studied how A
7
is distributed in such balls. He was able to estimate how much of A there is near
7
5 (n m)-planes. More precisely, assuming 0 m < s n and denoting
− ≤ ≤
8
0 X(x,V,α) = y Rn : dist(y x,V) < α y x ,
{ ∈ − | − |}
.
1 X(x,r,V,α) = X(x,V,α) B(x,r),
0 ∩
7 as x Rn, V G(n,m), r > 0, and 0 < α 1, he proved that there exists a
1 ∈ ∈ ≤
constant c = c(n,m,s,α) > 0 such that
:
v
i s A X(x,r,V,α)
X
limsup inf H ∩ c (1.2)
ar r↓0 V∈G(n,n−m) (cid:0) rs (cid:1) ≥
for s-almost every x A whenever A Rn is such that s(A) < . Here
G(nH,m) denotes the co∈llection of all m-d⊂imensional linear suHbspaces o∞f Rn, see
[12, 3.9]. Actually (1.2) is just a special case of Mattila’s result, as his theorem
§
can be applied also for more general cones, see [11, Theorem 3.3].
Date: January 31, 2017.
2000 Mathematics Subject Classification. Primary 28A75; Secondary 28A78, 28A80.
Key words and phrases. Conical density, porosity, Hausdorff dimension.
AK acknowledges the support of the Academy of Finland (project #114821).
1
2 ANTTI KA¨ENMA¨KIANDVILLE SUOMALA
In Theorem 2.5 we show that if A is as above, then it cannot be concentrated in
too small regions, not even inside the cones X(x,r,V,α). More precisely, denoting
H(x,θ) = y Rn : (y x) θ > 0 ,
{ ∈ − · }
H(x,θ,η) = y Rn : (y x) θ > η y x ,
{ ∈ − · | − |}
for x Rn, θ Sn 1, and 0 < η 1, we prove under the same assumptions as in
−
∈ ∈ ≤
(1.2) that there exists a constant c = c(n,m,s,α,η) > 0 such that
s A X(x,r,V,α) H(x,θ,η)
limsup inf H ∩ \ c
r 0 θ∈Sn−1 (cid:0) rs (cid:1) ≥
↓ V∈G(n,n−m)
for s-almost every x A. Here Sn 1 denotes the unit sphere of Rn. To help
−
H ∈
the geometric visualization, it might be helpful to take α and η close to 0 and
θ V Sn 1. Our method gives also a more elementary proof for (1.2) and it can
−
∈ ∩
also be used to obtain similar results for more general measures, see Theorem 2.7.
The nonsymmetric conical upper density theorem is essential in our application
to k-porous sets, that is, the sets with por > 0, see (1.5). The notation of poros-
k
ity, or 1-porosity using our terminology, has arisen from the study of dimensional
estimates related, for example, to the boundary behavior of quasiconformal map-
pings. See Koskela and Rohde [9], Martio and Vuorinen [10], Sarvas [15], Trocenko
[17], and V¨ais¨ala¨ [18]. The dimensional properties of 1-porous sets are well known.
Using a version of (1.2), Mattila showed that if porosity is close to its maximum
value 1, then the dimension cannot be much bigger than n 1. More precisely,
2 −
sup s > 0 : por (A) > ̺ and dim (A) > s for some A Rn n 1 (1.3)
{ 1 H ⊂ } −→ −
as̺ 1. Heredim referstotheHausdorffdimension. LaterSalli[14]generalized
→ 2 H
this result for the Minkowski dimension, and found the correct asymptotics. The
concept of 1-porosityhas also been generalized for measures, andit leads to similar
kindofdimensionbounds. SeeJ¨arvenpa¨¨aandJ¨arvenpa¨¨a[4]andreferencestherein.
Motivated by the fact that each V G(n,n 1) has maximal 1-porosity, we
∈ −
introduce a porosity condition which describes also sets whose dimension issmaller
than n 1. For any integer 0 < k n, x Rn, A Rn, and r > 0 we set
− ≤ ∈ ⊂
por (A,x,r) = sup ̺ : there are z ,...,z Rn such that
k { 1 k ∈
B(z ,̺r) B(x,r) A for every i, (1.4)
i
⊂ \
and (z x) (z x) = 0 for i = j .
i j
− · − 6 }
Here is the inner product. The k-porosity of A at a point x is defined to be
·
por (A,x) = liminfpor (A,x,r),
k k
r 0
↓
and the k-porosity of A is given by
por (A) = inf por (A,x). (1.5)
k k
x A
∈
NONSYMMETRIC CONICAL UPPER DENSITY AND k-POROSITY 3
This means that k-porous sets have holes in k orthogonal directions near each
of its points in every small scale. We shall now give a concrete example where
k-porosity occurs naturally. Suppose 0 < λ < 1 and let C R be the usual
2 λ ⊂
λ-Cantor set, see [12, 4.10]. It is clearly a 1-porous set with por (C ) 1 λ.
§ 1 λ ≈ 2 −
Mattila’s result (1.3) implies that dim (C ) 0 as por (C ) 1. Of course, we
H λ → 1 λ → 2
could obtain the same information just by calculating the Hausdorff dimension of
the self-similar set C and letting λ 0, but our aim was to provide the reader
λ
with an illustrative example. The se→ts C C R2 and C C [0,1] R3
λ λ λ λ
× ⊂ × × ⊂
are clearly 2-porous with por 1 λ. For these sets (1.3) does not give any
2 ≈ 2 −
reasonable dimension bound. However, it would be desirable to see, also in terms
of porosity, that dim (C C ) 0 and dim (C C [0,1]) 1 as λ 0.
H λ λ H λ λ
× → × × → →
This follows as an immediate application of Theorem 3.2. Using our nonsymmetric
conical upper density theorem, we show that
sup s > 0 : por (A) > ̺ and dim (A) > s for some A Rn n k
{ k H ⊂ } −→ −
as ̺ 1. Observe also that in the proof of Theorem 3.2 the orthogonality in (1.4)
→ 2
plays no rˆole and we may replace it by an assumption of a uniform lower bound
for the angles between z x and the (k 1)-plane spanned by vectors z x,
i j
− − −
i = j.
6
Let us now discuss the situation when porosity is small. It is well known (for
example, see [10]) that if A Rn with por (A,x,r) ̺ > 0 for all x A and
⊂ 1 ≥ ∈
0 < r < r , then
0
dim (A) < n c̺n, (1.6)
M
−
where c > 0 depends only on n, and dim refers to the Minkowski dimension, see
M
[12, 5.3]. It might be possible to get a better estimate if por is replaced by por
§ 1 k
for some k > 1, but this condition does not feel very natural if the size of the holes
is small. However, if V G(n,m) is fixed and the condition por (A,x,r) ̺ is
∈ 1 ≥
replaced by
sup ̺ : B(z,̺r) B(x,r) A for some z V + x ̺,
′ ′
⊂ \ ∈ { } ≥
then n in (1.6) can be replaced by m, see Theorem 4.3. This is a rather immediate
(cid:8) (cid:9)
consequence of (1.6), but our main point is to give a simple proof for (1.6) using
iterated function systems.
Acknowledgement. The authors are indebted to Professor Pertti Mattila for his
valuable comments for the manuscript. The authors thank also Esa Ja¨rvenpa¨¨a,
Maarit J¨arvenpa¨¨a, Pekka Koskela, Tomi Nieminen, Kai Rajala and Eero Saksman
for useful discussions during the preparation of this article.
2. Nonsymmetric conical upper density
We shall first prove a density theorem for nonsymmetric regions and then prove
our main theorem by using a similar argument on (n m)-planes. The proofs rely
−
on the following geometric fact.
4 ANTTI KA¨ENMA¨KIANDVILLE SUOMALA
z w
Figure A. All points lying on the gray region form a large angle
with points z and w.
Lemma 2.1. For given 0 < β < π, there is q = q(n,β) N such that in any set
of q points in Rn, there are always three points which det∈ermine an angle between
β and π.
Remark 2.2. Erd˝os and Fu¨redi [1] have shown that for the smallest possible choice
of q it holds that
2(π/(π β))n−1 q(n,β) 2(4π/(π β))n−1 +1.
− −
≤ ≤
For the convenience of the reader we shall give below a different proof which
establishes the existence of some such q. The estimate that we get here for q is,
however, quite bad compared to the best possible one.
Proof. Let A be a set of points in Rn so that all angles formed by its points are
less than β. Let us fix 0 < η < 1 and cover Rn 0 by cones C = H(0,θ ,η),
i i
i 1,2,...,k , where the constant k = k(n,η) \N{d}epends only on n and η. To
∈ { } ∈
visualize the situation, note that if β is close to π, then η is close to 1 and cones C
i
are very narrow. To simplify the notation, we denote C = C + y for y Rn.
i,y i
For any index i i i , where j N and i 1,2,...,k for{1} m ∈j, we
1 2 j m
··· ∈ ∈ { } ≤ ≤
define sets A in the following way: We begin by fixing x A and setting
i1i2···ij ∈
A = A C for 1 i k. If A has been defined, we choose y A
i ∩ i,x ≤ ≤ i1i2···ij ∈ i1i2···ij
and define A = A C for 1 l k (if A is empty, then so
i1i2···ijl i1i2···ij ∩ l,y ≤ ≤ i1i2···ij
is A ). We refer to y as the corner of A . It follows directly from the
i1i2 ijl i1i2 ijl
··· ···
definition of the sets A that
i1i2 ij
···
k
cardA 1+ cardA .
i1i2···ij ≤ i1i2···ijl
l=1
X
Iterating this, we get
k k
cardA kj + cardA . (2.1)
≤ i1i2···ikl
Xj=0 i1Xi2···ikXl=1
The main point of the proof is the observation that if η = η(β) is chosen to
be close enough to 1 in the beginning, then the following is true: If z and w
are the corners of A and A , respectively, and if z C , then
i1i2···ij i1i2···ijij+1···im ∈ im,w
A C C = . See Figure A. It follows by induction from the above fact
∩ ij,z ∩ im,w ∅
NONSYMMETRIC CONICAL UPPER DENSITY AND k-POROSITY 5
w
r
√s2 1r
−
sr z θ
y
Figure B. Illustration for the proof of Lemma 2.3.
that for given A we have
i1i2 ij
···
card l : A = k j.
{ i1i2···ijl 6 ∅} ≤ −
In particular, A = for any choice of i i ...i . Combined with (2.1),
i1i2···ik+1 ∅ 1 2 k+1
this gives cardA k kj. This number depends only on k = k(n,β) and the
≤ j=0 (cid:3)
claim follows.
P
For 0 < η 1 we define
≤
η2 +4
t(η) = ,
s η2
1
γ(η) = .
t(η)
Notice that t(η) 2 and η/√5 γ(η) η/2.
≥ ≤ ≤
Lemma 2.3. Suppose that y Rn, θ Sn 1, 0 < η 1, t = t(η), and γ = γ(η).
−
If z Rn B(y,tr) H(y,θ,∈γ) , then∈ ≤
∈ \ ∪
B(z,r) H(y,θ,η) = .
(cid:0) (cid:1)
∩ ∅
Proof. Take w Rn such that it maximizes (w y) θ/ w y in the closure
∈ − · | − |
of B(z,r). It suffices to prove that (w y) θ/ w y < η, see Figure B. It
− · | − |
is straightforward to check that η√s2 1 1 + γs when s t. Denoting now
− ≥ ≥
s = y z /r, we have s t > 1 and thus
| − | ≥
(w y) θ < r +γ y z = (1+γs)r
− · | − |
η√s2 1r = η w y ,
≤ − | − |
(cid:3)
which finishes the proof.
Theorem 2.4. Suppose 0 < η 1 and 0 < s n. Then there is a constant
≤ ≤
c = c(n,s,η) > 0 such that
s A B(x,r) H(x,θ,η)
limsup inf H ∩ \ c
r↓0 θ∈Sn−1 (cid:0) rs (cid:1) ≥
for s almost every x A whenever A Rn with s(A) < .
H ∈ ⊂ H ∞
6 ANTTI KA¨ENMA¨KIANDVILLE SUOMALA
Proof. Takec > 0andassumethatthereexists aBorelsetB Rn with s(B) > 0
⊂ H
such that for each x B and 0 < r < r there is θ Sn 1 for which
0 −
∈ ∈
s B B(x,r) H(x,θ,η) < crs. (2.2)
H ∩ \
It suffices to find a positive lower bound for c in terms of n, s, and η.
(cid:0) (cid:1)
Using (1.1), and replacing B by a suitable subset if necessary, we may assume
that
s B B(x,r) < 2s+1rs (2.3)
H ∩
for all 0 < r < r and x B. Moreover, using the lower estimate of (1.1), we find
0 (cid:0) (cid:1)
∈
0 < r < r /3 and x B such that
0
∈
s B B(x,r) > 1rs. (2.4)
H ∩ 2
Set t = t(η), γ = γ(η), and t(cid:0)ake 0 < δ <(cid:1) 1. Let us fix β < π such that the
opening angle of H(x,θ,γ) is smaller than β, and let q = q(n,β) be as in Lemma
2.1. We may cover the set B B(x,r) by 4nδ n balls of radius δr with centers in
−
∩
B. Using (2.4), we notice that there exists x B B(x,r) such that
1
∈ ∩
s B B(x ,δr) > 4 nδn2 1rs.
1 − −
H ∩
The set B B(x,r) B(x1(cid:0),tδr) can also(cid:1)be covered by 4nδ−n balls of radius δr
∩ \
with centers in B. Whence, using (2.3) and (2.4),
s B B(x,r) B(x ,tδr) > (1 2s+1tsδs)rs.
H ∩ \ 1 2 −
If 12 −2s+1tsδs > 0,(cid:0)we find x2 ∈ B ∩B(x,r)(cid:1)\B(x1,tδr) for which
s B B(x ,δr) > 4 nδn(1 2s+1tsδs)rs.
H ∩ 2 − 2 −
Choosing δ = δ(n,s,η(cid:0)) > 0 small en(cid:1)ough and continuing in this manner, we find
q points x ,...,x B B(x,r) with x x tδr for i = j, such that for each
1 q i j
∈ ∩ | − | ≥ 6
i 1,...,q we have
∈ { }
s B B(x ,δr) > 4 nδn 1 (q 1)2s+1tsδs rs
H ∩ i − 2 − − (2.5)
=: c(n,s,η)(3r)s,
(cid:0) (cid:1) (cid:0) (cid:1)
where c(n,s,η) > 0.
According to Lemma 2.1, we may choose three points y,y ,y from the set
1 2
x ,...,x such that for each θ Sn 1 there is i 1,2 for which y Rn
1 q − i
{ } ∈ ∈ { } ∈ \
B(y,tδr) H(y,θ,γ) . We obtain, using Lemma 2.3, that for each θ Sn 1 there
−
∪ ∈
is i 1,2 such that
(cid:0) ∈ { } (cid:1)
B(y ,δr) B y,2(1+δ)r H(y,θ,η).
i
⊂ \
Thus, applying (2.5), we have (cid:0) (cid:1)
s B B(y,3r) H(y,θ,η) > c(n,s,η)(3r)s
H ∩ \
for all θ Sn−1. Re(cid:0)calling (2.2), we conclud(cid:1)e that c c(n,s,η). The proof is
∈ ≥ (cid:3)
finished.
NONSYMMETRIC CONICAL UPPER DENSITY AND k-POROSITY 7
Theorem 2.5. Suppose 0 < α,η 1 and 0 m < s n. Then there is a
≤ ≤ ≤
constant c = c(n,m,s,α,η) > 0 such that
s A X(x,r,V,α) H(x,θ,η)
limsup inf H ∩ \ c
r 0 θ∈Sn−1 (cid:0) rs (cid:1) ≥
↓ V∈G(n,n−m)
for s almost every x A whenever A Rn with s(A) < .
H ∈ ⊂ H ∞
Proof. For any V,W G(n,n m), we set d(V,W) = sup dist(x,W).
With this metric G(n,∈n m) is a−compact metric space, see Saxl∈liV[∩1S3n].−1Defining for
−
each V G(n,n m) a set W : d(V,W) < α/2 we notice that a finite number of
∈ − { }
these sets is still a cover. We assume that the sets assigned to the planes V ,...,V ,
1 l
where l = l(n,m,α), cover G(n,n m). For any W, it holds that d(V ,W) < α/2
i
−
with some i 1,...,l . This implies X(0,V ,α/2) X(0,W,α). Thus, for each
i
∈ { } ⊂
W G(n,n m), there is i such that
∈ −
X(x,r,W,α) X(x,r,V ,α/2) (2.6)
i
⊃
for all r > 0 and x Rn. We shall prove that if A Rn with s(A) < , then
∈ ⊂ H ∞
s A X(x,r,V ,α/2) H(x,θ,η)
i
limsup inf H ∩ \ c(n,m,s,α,η)
r 0 θ∈Sn−1 (cid:0) rs (cid:1) ≥
↓ i∈{1,...,l}
for s almost every x A from which the claim follows easily by using (2.6).
THake c > 0 and assu∈me that there is a Borel set B Rn with s(B) > 0 such
⊂ H
that for each x B and 0 < r < r there are i and θ Sn 1 for which
0 −
∈ ∈
s B X(x,r,V ,α/2) H(x,θ,η) < crs.
i
H ∩ \
According to (1.1) we(cid:0)may assume that (cid:1)
s B B(x,r) < 2s+1rs (2.7)
H ∩
for all 0 < r < r0 and x B. U(cid:0) sing the low(cid:1)er estimate of (1.1), we find 0 < r <
∈
r /3 and x B such that
0
∈
s B B(x,r) > 1rs. (2.8)
H ∩ 2
Next we define (cid:0) (cid:1)
B = z B : s B X(z,3r,V ,α/2) H(z,θ,η) < c(3r)s
i i
∈ H ∩ \ (2.9)
for some θ Sn 1 .
(cid:8) (cid:0) (cid:1) −
∈
Since l B = B, we infer from (2.8) that there is i 1,...,l fo(cid:9)r which
i=1 i 0 ∈ { }
S Hs Bi0 ∩B(x,r) > 2−1l−1rs.
Let t = max 5/α,t(η) , ch(cid:0)oose q = (n,η(cid:1)) as in the proof of Theorem 2.4, and
{ }
define 0 < ε < 1 so that
4 m2 1l 1εm (q 1)2s+1tsεs = 4 m 1l 1εm, (2.10)
− − − − − −
− −
8 ANTTI KA¨ENMA¨KIANDVILLE SUOMALA
z
y
w
4εr/α
V + x
{ }
x
r
Figure C. Illustration for the proof of Theorem 2.5.
recall that s > m so that this is possible. Since the set (V + x ) B(x,r) may
i⊥0 { } ∩
be covered by 4mε m balls of radius εr, there exists y (V + x ) B(x,r) such
− ∈ i⊥0 { } ∩
that
s B B(x,r) P 1(B(y,εr)) > 4 m2 1l 1εmrs. (2.11)
H i0 ∩ ∩ V−i⊥0 − − −
We now argue as(cid:0)in the proof of Theorem 2.4(cid:1)above. We first observe that the
slice S = B B(x,r) P 1 B(y,εr) may be covered by c 1εm n balls of radius
i0 ∩ ∩ V−i⊥0 −1 −
εr for a constant c = c (n,m) > 0. Then we use (2.11), (2.7), and (2.10) to find
1 1 (cid:0) (cid:1)
points x ,...,x S such that x x tεr whenever i = j and
1 q i j
{ } ∈ | − | ≥ 6
s S B(x ,εr) > c εn m 4 m2 1l 1εmrs (q 1)2s+1tsεsrs
i 1 − − − −
H ∩ − − (2.12)
= c (3r)s
(cid:0) (cid:1) 2 (cid:0) (cid:1)
for all i. Here c = c (n,m,s,α,η) = c 3 s4 m 1l 1εm. Now the same geo-
2 2 1 − − − −
metric argument as in the proof of Theorem 2.4 implies that there is a point
z x ,...,x such that for each θ Sn 1 we may find w x ,...,x z
1 q − 0 q
∈ { } ∈ ∈ { }\{ }
so that
B(w,εr) B z,(2+ε)r H(z,θ,η) B(z,4εr/α) .
⊂ \ ∩
Since also
(cid:0) (cid:1) (cid:0) (cid:1)
P 1 B(y,εr) B(z,3r) B(z,4εr/α) X(z,3r,V ,α/2),
V−i⊥0 ∩ \ ⊂ i0
see Figure C, w(cid:0)e get (cid:1)
inf s B X(z,3r,V ,α/2) H(z,θ,η) c (3r)s.
θ Sn−1H ∩ i0 \ ≥ 2
∈
by(2.12). Nowz B (cid:0)andweconclude, using (2.9),that(cid:1)c c = c (n,m,s,α,η).
∈ i0 ≥ 2 2 (cid:3)
This completes the proof.
NONSYMMETRIC CONICAL UPPER DENSITY AND k-POROSITY 9
Remark 2.6. Inspecting the proofs, one can read explicit expressions for the con-
stants in Theorems 2.4 and 2.5. In Theorem 2.4, one gets c 2c1/( sηn−1) and
−
in Theorem 2.5, one obtains c αc3/(s m)2c2/((m s)ηn−1). Th≥e constants 0 <
− −
≥
c ,c ,c < here depend only on n. The estimates obtained in this way are
1 2 3
∞
probably rather far from being optimal, although the best values are not known.
Our method can be applied also in a more general setting. A similar proof as
above gives the following result. If µ is a measure on Rn, h: (0,r ) (0, ), and
0
→ ∞
x Rn, we define D(µ,x) and D(µ,x) as the lower and upper limits, respectively,
∈
of the ratio µ B(x,r) /h(r) as r 0.
↓
Theorem 2.7(cid:0). Suppo(cid:1)se 0 m < n and h: (0,r ) (0, ) is a function with
0
≤ → ∞
h(εr)
0 uniformly for all 0 < r < r (2.13)
εmh(r) −→ 0
as ε 0. Let µ be a measure on Rn with D(µ,x) < for µ-almost all x Rn.
↓ ∞ ∈
For every 0 < α,η 1, there is a constant c = c(n,m,h,α,η) > 0 such that
≤
µ X(x,r,V,α) H(x,θ,η)
limsup inf \ cD(µ,x)
r 0 θ∈Sn−1 (cid:0) h(r) (cid:1) ≥
↓ V∈G(n,n−m)
for µ-almost every x Rn.
∈
Let us make few comments related to the above theorem. Suppose that h
fulfills condition (2.13). Let be the generalized Hausdorff measure which is
h
H
constructedusinghasagaugefunction, see[12, 4.9]. Ifµ = ,where (A) <
h A h
§ H | H
, then D(µ,x) < for µ-almost every x Rn, and thus Theorem 2.7 can be
∞ ∞ ∈
applied.
There are many natural gauge functions, such as h(r) = rslog(1/r) where m <
s < n, which satisfy (2.13). However, some interesting cases, such as h(r) =
rm/log(1/r), are not covered by this condition.
It seems to be unknown whether a similar result as Theorem 2.7 holds if one
replaces the condition D(µ,x) < by D(µ,x) < . The most interesting ex-
∞ ∞
ample falling into this category is obtained when µ = s and h(r) = rs, where
A
P |
s(A) < and m < s < n. Here s denotes the s-dimensional packing measure,
P ∞ P
see [12, 5.10]. See also Suomala [16] for related theorems.
§
3. Sets with large k-porosity
Mattila [11] proved Theorem 2.5 in the case m = n 1. Using this, he obtained
−
the desired dimension bounds for 1-porous sets, see (1.3). Our result for k-porous
sets follows applying a similar argument.
10 ANTTI KA¨ENMA¨KIANDVILLE SUOMALA
For √2 1 < ̺ < 1 we define
− 2
1
t(̺) = ,
√1 2̺
−
1 ̺ ̺2 +2̺ 1
δ(̺) = − − − .
√1 2̺
p−
Notice that δ(̺) 0 as ̺ 1.
→ → 2
Lemma 3.1. Suppose x Rn, r > 0, √2 1 < ̺ < 1, t = t(̺), and δ = δ(̺). If
z Rn x is such that∈B(z,̺tr) B(x,−tr), then 2
∈ \{ } ⊂
H(x+δrθ,θ) B(x,r) B(z,̺tr),
∩ ⊂
where θ = (z x)/ z x .
− | − |
Proof. Tosimplifythenotation,weassumer = 1, x = 0, andθ = e = (1,0,...,0).
1
This will not affect the generality. Let y B(0,1) B(z,̺t). We have to show
∈ \
that
y / H(x+δθ,θ). (3.1)
∈
By the Pythagorean Theorem we have
z y = z y 2 y y 2 (̺t)2 1.
1 1
| − | | − | −| − | ≥ −
Using this, we obtain p p
y = z z y t ̺t (̺t)2 1 = δ,
1 1
| |−| − | ≤ − − −
(cid:3)
which implies (3.1). p
Theorem 3.2. Suppose 0 < k n. Then
≤
sup s > 0 : por (A) > ̺ and dim (A) > s for some A Rn n k
{ k H ⊂ } −→ −
as ̺ 1.
→ 2
Proof. Assume on the contrary that there exists s > n k such that for each
−
√2 1 < ̺ < 1 there is a set A for which dim (A ) > s and por (A ) > ̺. Take
− 2 ̺ H ̺ k ̺
√2 1 < ̺ < 1 and such a set A . Now A has a subset B for which dim (B) > s
− 2 ̺ ̺ H
and por (B,x,r) > ̺ for all x B and 0 < r < r with some r > 0. Clearly also
k ∈ 0 0
the closure of B satisfies these conditions. Thus there is a closed set F B (for
⊂
example, use [2, Theorem 5.4]) such that 0 < s(F) < and
H ∞
por (F,x,r) > ̺ for all x F and 0 < r < r .
k ∈ 0
Therefore, for any x F and 0 < r < r /t, there are z ,...,z Rn such that
0 1 k
∈ ∈
B(z ,̺tr) B(x,tr) F for i = 1,...,k, and (z x) (z x) = 0 for i = j. Put
i i j
⊂ \ − · − 6
θ = (z x)/ z x . Applying now Lemma 3.1 we have H(x+δrθ ,θ ) B(x,r)
i i i i i
− | − | ∩ ⊂